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I recently saw in another post that $(\ln{f})'=\frac{f'}{f}$

where $f=f(x)$

From which it follows that $\int\frac{f'}{f}\mathrm dx=\int(\ln{f})'\mathrm dx=\ln{f}+C$

What about integrating the inverse of this?

I.e. what about $\int\frac{f}{f'}\mathrm dx=\int\frac{1}{(\ln{f})'}\mathrm dx=?$

ben
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3 Answers3

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$$\int\frac{f'}{f}dx=\int(\ln{f})'dx=\ln{f}+C$$ is actually just a substitution :$u=f(x) \,;\, du =f'(x) dx$.

About the second question:

$$\int\frac{f}{f'}dx=\int\frac{1}{(\ln{f})'}dx$$

Suppose that such formula exists. Let $g(x)$ be any function and let $f(x)=e^{g(x)}$.

Then the above formula would yield a general formula for

$$\int \frac{1}{g'(x)} dx \,.$$

Conversely if a formula for $\int \frac{1}{g'(x)} dx \,.$ exists then you can get your formula by defining $g(x):= \ln |f(x) | \,.$

The question you ask is equivalent to the existence of a formula for

$$\int \frac{1}{g'(x)} dx \,.$$

I highly doubt that this is true, but couldn't find the right transcendent function, I am sure someone smarter will ;)

N. S.
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We give an explicit example that completes the argument of @user9176. Note first that $\dfrac{e^x}{x}$ does not have an elementary antiderivative. For a proof, see this. But in the notation of @user9176, $$\frac{e^x}{x}=\frac{1}{g'(x)},$$ where $g'(x)=xe^{-x}$. Since $\int xe^{-x}\,dx=-(xe^{-x}+e^{-x})+C$, the function $g(x)$ is an elementary function, and therefore so is $f(x)$, where $f(x)=e^{g(x)}$.

André Nicolas
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  • For references to non-elementary integrals, see http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral/2334#2334 – lhf Oct 30 '11 at 19:38
  • @lhf: There was a link there, but I couldn't figure out how to make it work. Just now realized the http was missing. – André Nicolas Oct 30 '11 at 20:42
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I wonder if you know about change of variables for an indefinite integral. You can read about it on the linked website, but the main idea is that $$ \int G'(f(x))f'(x)dx = G(f(x))+c. $$

In your case, $G(f) = \log f$ so $\displaystyle{G'(f) = \frac1f}$ and hence $\displaystyle{\frac{f'}f = G'(f)f'}$.

On the other hand, there is no such a function $G$ that $\displaystyle{G(f)f' = \frac{f}{f'}}$ for any function $f$ which is, say differentiable. The naive explanation is that if such function would exist then $\displaystyle{G(f) = \frac f{f'^2}}$ which is impossible since the LHS depends only on $f$ while RHS depends on both $f$ and $f'$.

SBF
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